tidyr

I have a dataframe in the following format: id | name | logs ---+--------------------+----------------------------------------- 84 | "zibaroo" | "C47931038" 12 | "fabien kelyarsky" | c("C47331040", "B19412225", "B18511449") 96 | "mitra lutsko" | c("F19712226", "A18311450") 34 | "PaulSandoz" | "A47431044" 65 | "BeamVision" | "D47531045" As you see the column "logs" includes vectors of strings in each cell. Is there an efficient way to convert the data frame to the long format (one observation per row) without the intermediary step of separating "logs" into several columns? This is important

I have a data frame that looks like this: a b 1 x 8 2 x 6 3 y 3 4 y 4 5 z 5 6 z 6 and I want to turn it into this: x y z 1 8 3 5 2 6 4 6 But calling library(tidyr) df <- data.frame( a = c("x", "x", "y", "y", "z", "z"), b = c(8, 6, 3, 4, 5, 6) ) df %>% spread(a, b) returns x y z 1 8 NA NA 2 6 NA NA 3 NA 3 NA 4 NA 4 NA 5 NA NA 5 6 NA NA 6 What am I doing wrong? While I'm aware you're after tidyr , base has a solution in this case: unstack(df, b~a) It's also a little bit faster: Unit: microseconds expr min lq mean median uq max neval df %>% spread(a, b) 657.699 679.508 717.7725 690.484 724.9795

This question already has an answer here: Reshaping data in R with “login” “logout” times 5 answers My data looks like this: I am trying to make it look like this: I would like to do this in tidyverse using %>%-chaining. df <- structure(list(id = c(2L, 2L, 4L, 5L, 5L, 5L, 5L), start_end = structure(c(2L, 1L, 2L, 2L, 1L, 2L, 1L), .Label = c("end", "start"), class = "factor"), date = structure(c(6L, 7L, 3L, 8L, 9L, 10L, 11L), .Label = c("1979-01-03", "1979-06-21", "1979-07-18", "1989-09-12", "1991-01-04", "1994-05-01", "1996-11-04", "2005-02-01", "2009-09-17", "2010-10-01", "2012-10-06" ), class

I'm doing a series of things in dplyr , tidyr , so would like to keep with a piped solution if possible. I have a list with uneven numbers of elements in each component: lolz <- list(a = c(2,4,5,2,3), b = c(3,3,2), c=c(1,1,2,4,5,3,3), d=c(1,2,3,1), e=c(5,4,2,2)) lolz $a [1] 2 4 5 2 3 $b [1] 3 3 2 $c [1] 1 1 2 4 5 3 3 $d [1] 1 2 3 1 $e [1] 5 4 2 2 I am wondering if there's a neat one liner to fill up each element with NAs such that they all are of the same length as the element with the maximum items: I have a 2 liner: lolz %>% lapply(length) %>% unlist %>% max -> mymax lolz %>% lapply(function

Suppose I have some count data that looks like this: library(tidyr) library(dplyr) X.raw <- data.frame( x = as.factor(c("A", "A", "A", "B", "B", "B")), y = as.factor(c("i", "ii", "ii", "i", "i", "i")), z = 1:6) X.raw # x y z # 1 A i 1 # 2 A ii 2 # 3 A ii 3 # 4 B i 4 # 5 B i 5 # 6 B i 6 I'd like to tidy and summarise like this: X.tidy <- X.raw %>% group_by(x,y) %>% summarise(count=sum(z)) X.tidy # Source: local data frame [3 x 3] # Groups: x # # x y count # 1 A i 1 # 2 A ii 5 # 3 B i 15 I know that for x=="B" and y=="ii" we have observed count of zero, rather than a missing value. i.e. the

I have a simple question, which I can't figure out. I have a dataframe with two factors ( distance ) and years ( years ). I would like to complete all years values for every factor by 0. i.e. from this: distance years area 1 NPR 3 10 2 NPR 4 20 3 NPR 7 30 4 100 1 40 5 100 5 50 6 100 6 60 get this: distance years area 1 NPR 1 0 2 NPR 2 0 3 NPR 3 10 4 NPR 4 20 5 NPR 5 0 6 NPR 6 0 7 NPR 7 30 8 100 1 40 9 100 2 0 10 100 3 0 11 100 4 0 12 100 5 50 13 100 6 60 14 100 7 0 I tried to apply expand() function: library(tidyr) library(dplyr, warn.conflicts = FALSE) expand(df, years = 1:7) but this just